In week 17 of precalc, we went over Sine law and Cosine law. We use these set of rules to find angles and sides of NON RIGHT triangles (triangles without a 90° angle).

Sine law has the formula SinA/a=SinB/b=SinC/c OR: a/SinB=b/SinB=c/SinC. For reference, the capitals are the angles and lowercase letters are sides. The thing with sine law is that you need to know which version to use for what you’re finding. Basically, when you are finding the angle, put angles (SinA,B,C) on the top and the sides (abc) on the bottom and vice versa for when you are finding sides.

The sine rule is useful when you have two angles and a side. The sine rule helps to find the remaining sides and angles. It’s also helpful when you are given two sides and a non-included angle. The sine rule can help to find the other angles and the third side, though this can sometimes lead to the ambiguous case where two different triangles satisfy the given conditions.

Cosine law has the formula b^2=a^+b^2(2(ac))cosB, or more simplified:

The cosine law, similarly to Pythagoras (A^2+B^2=C^2), you can interchange the variables, like to find B, just replace the values accordingly.

The cosine rule is useful when you have two sides and an included angle. cosine rule helps to find the third side. It also works when you know all three sides of the triangle. the cosine rule helps to find any of the angles.

Now let’s go through example to show the use of the cosine rule

Problem: You have a triangle with sides a=7units, $=5$ units, and the included angle C=60 degrees. Find the length of the side c. Since we know two sides and the included angle we use the cosine rule: c^2=a^2+b^2−2abcos(C). Substituting the given values: c^2=7^2+5^2−2⋅7⋅5⋅cos(60). cos 60 is just .5 so c^2=49+25−2⋅7⋅5⋅0.5. Simplifying it turns the equation into c^2=49+25−35 and then c^2=39 so our answer is c=√39.

In week 16 of precalc 11 we went over more trig. We learned how SOH CAH TOA works on a chart. SOH CAH TOA can be translated to SYR CXR TYX (R being the rotation angle). This is how you find triangles on a graph. Most problems we work with involve finding R with Pythagoras theorem, then using SYR CXR TYX to find an angle inside the triangle. We also learned about an important rule called CAST. basically it goes counter clockwise from Q4, being C, Q1 being A, Q2 being S and Q3 being T. The acronym is for Cosine, All, Sine and Tangent. In their respective quadrants, they are positive and they are negative in all other ones.

The problems we will work with is seeing if something is positive or negative. for example we will have something like “is sin 305 positive or negative?” then we will find that 305 is in Q4 or (or the C quadrant) and since it’s sine, it is negative. It’s negative because since sin = y/r, the Y in Q4 is negative, making the fraction negative, making sin a negative number.

Another thing we learned about is special triangles. Basically, 30, 45 and 60 degrees (and rotation angles with those as reference angles) are important to remember because you don’t even need a calculator to work with them. We can memorize the exact values of a triangle with these degrees. Basically we memorized these two triangles and their values.

In week 15 of precalc we started going into trigonometry. This meant stuff like rotation angles and triangles. It’s usually finding the length of one side or the angle of two sides of a triangle using Sine, Cosine and Tangent. To review on how it works, there’s an abbreviation called SOH CAH TOA. Basically it means Sine = Opposite/Hypotenuse (SOH), Cosine= Adjacent/Hypotenuse (CAH) and Tangent = Opposite/Adjacent (TOA).

Now when we have a point on a graph it has its own rotation angle to what is known as the starting arm. In Trig, you try to find the angle between the two arms. You can have more than one arm on a graph, and they can have the same REFERENCE ANGLE but have different ROTATION ANGLES. The reference angle is in the first quadrant and can be used to find the rotation angle of other arms, for example if you have an arm in the first and third quadrants, you can add 180 to the rotation angle and that will be the rotation angle of the arm in Q3, for example the Rotation angle of 45 in Q3 is 225 because 45+180=225. For another example, the reference angle of 348 is 12 because 360-348=12.

As another way to explain it, you either add or subtract the reference angle from the closest horizontal arm, depending on the position of the terminal arm to find the rotation angle.

In week 12 of precalc 11 we went over the basics of non-permissible values in fractions. Basically, if the value of the denominator (bottom of the fraction) is zero, then it doesn’t work. This is because you can’t divide anything by 0, like imagine taking groups of 0 out of something, it just doesn’t work.

Now the fractions we’ve been working with usually have an X value and an integer on the bottom. We have to find a way to make it 0 with our given numbers (We have to find out how to make it 0 so we can avoid making it 0).

Basically if we have X-5 on the bottom, X can NOT equal 5 because 5-5 is zero.

To make it more complicated we can have a coefficient on X, so we could have it be 2X-5 on the bottom. To make the bottom be zero we will have to first divide the 2x by two and so we must also divide -5 by 2, resulting in X-5/2 on the bottom. Then X will have to be NOT equal to 5/2, because 5/2-5/2=0

To make it even more complicated we can have binomials such as (2X-5) (X+7) on the bottom. To find a binomial’s non permissible values, you will have to solve both the brackets, the logic behind this is that the brackets multiply, so if just one of them equal 0, then it will multiply the other by 0, resulting in 0. Nonetheless you must still solve both brackets because either bracket could multiply the other or something like that. so it will be X≠5/2 and/or -7.

Long story short, if you can make the bottom of a fraction zero, the numbers you used will be known as non-permissible.

In week 10 of preacalc, we learned how to convert quadratic functions from general form (AX^2+BX+C=0) to standard form (Y=A(X−C)^2+D). We can do it by completing the square.

Here are the steps to convert a quadratic function from general form to standard form:

1. Make sure the coefficient of x^2 is 1: If the coefficient of x^2 is not $1$, factor it out. For example, if your quadratic function is 3x^2+6x−9, divide each term by 3 to get 2( x^2+2x−3).
2. Complete the square: To complete the square, you need to add and subtract (b/2)^2 inside the parentheses. In other words, take half of the coefficient of $x$, square it, and add/subtract it. So for ax^2+bx, you would add (b/2)^2 and subtract it to the expression. In our example, x^2+2x, half of $b$ is $1$, so we add and subtract x^2+2x+1−1
3. Rearrange the expression: Rearrange the expression so that the part you completed the square on is a perfect square trinomial. (x2+2x+1)−1
4. Combine like terms: Combine any like terms in the expression. 2(x+1)−1
5. Determine the value of $D$: In the standard form, D represents the y-coordinate of the vertex. In this case, D=−1.
6. Determine the value of $C$: The x-coordinate of the vertex. In this case, C=−1.
7. So the standard form of the quadratic function x^2+2x−3 is 2(x+1)−1.“>

In week 9 of Precalc, we dove into quadratic functions, we learned about the standard form of quadratic functions and this is the formula:

Y=A(X−C)^2+D

the A and D are the x and y of a point, usually introduced in a question, and the X and Y values will be the vertex, also introduced in the same question. The way this formula works is you have to find the A value by replacing ACXY by their given units, then rewrite it as Y=A(X−C)^2+D with the X and Y as the variables X and Y, this is how you graph it.

The way to graph a quadratic function is simple when you know what every unit does.

The A unit is the stretch, it will effect the parabola, making it wider or thinner.

The (X−C)^2 tells you where the vertex will be on the X axis. The way it works is weird because negative actually makes it move up by positive C. this means when you see (X−6)^2 for example, the vertex will be on X6, and so will the axle of symmetry.

The +D will basically move the vertex vertically. whatever D is, is where your vertex is on the Y axis.

Putting it all together, this is what the function Y=3(X-5)^2+2 will look like